Given the right triangles ABC and ABD, what is the length of segment BC, in units? [asy]
size(150);
pair A, B, C, D, X;
A=(0,0);
B=(0,12);
C=(-16,0);
D=(-35,0);
draw(A--B--D--A);
draw(B--C);
draw((0,1.5)--(-1.5,1.5)--(-1.5,0));
label("$37$", (B+D)/2, NW);
label("$19$", (C+D)/2, S);
label("$16$", (A+C)/2, S);
label("A", A, SE);
label("B", B, NE);
label("D", D, SW);
label("C", C, S);
[/asy]
Answer: Looking at triangle $ABD$, we can calculate $AB$ through the Pythagorean Theorem. \begin{align*}
AB&=\sqrt{37^2-(19+16)^2}\\
&=\sqrt{37^2-35^2}\\
&=\sqrt{(37+35)(37-35)}\\
&=\sqrt{72 \cdot 2}\\
&=\sqrt{36 \cdot 4}=6 \cdot 2=12\end{align*} Using Pythagorean theorem again to find $BC$, we get \begin{align*}
BC&=\sqrt{12^2+16^2}\\
&=\sqrt{4^2(3^2+4^2)}\\
&=4\sqrt{9+16}\\
&=4\sqrt{25}=4 \cdot 5=\boxed{20} \text{ units}.\end{align*}